3.292 \(\int \frac {1}{(a \cot (x)+b \csc (x))^5} \, dx\)

Optimal. Leaf size=100 \[ -\frac {4 b}{a^5 (a \cos (x)+b)}-\frac {\log (a \cos (x)+b)}{a^5}+\frac {\left (a^2-b^2\right )^2}{4 a^5 (a \cos (x)+b)^4}+\frac {4 b \left (a^2-b^2\right )}{3 a^5 (a \cos (x)+b)^3}-\frac {a^2-3 b^2}{a^5 (a \cos (x)+b)^2} \]

[Out]

1/4*(a^2-b^2)^2/a^5/(b+a*cos(x))^4+4/3*b*(a^2-b^2)/a^5/(b+a*cos(x))^3+(-a^2+3*b^2)/a^5/(b+a*cos(x))^2-4*b/a^5/
(b+a*cos(x))-ln(b+a*cos(x))/a^5

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Rubi [A]  time = 0.12, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4392, 2668, 697} \[ \frac {\left (a^2-b^2\right )^2}{4 a^5 (a \cos (x)+b)^4}+\frac {4 b \left (a^2-b^2\right )}{3 a^5 (a \cos (x)+b)^3}-\frac {a^2-3 b^2}{a^5 (a \cos (x)+b)^2}-\frac {4 b}{a^5 (a \cos (x)+b)}-\frac {\log (a \cos (x)+b)}{a^5} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cot[x] + b*Csc[x])^(-5),x]

[Out]

(a^2 - b^2)^2/(4*a^5*(b + a*Cos[x])^4) + (4*b*(a^2 - b^2))/(3*a^5*(b + a*Cos[x])^3) - (a^2 - 3*b^2)/(a^5*(b +
a*Cos[x])^2) - (4*b)/(a^5*(b + a*Cos[x])) - Log[b + a*Cos[x]]/a^5

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {1}{(a \cot (x)+b \csc (x))^5} \, dx &=\int \frac {\sin ^5(x)}{(b+a \cos (x))^5} \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a^2-x^2\right )^2}{(b+x)^5} \, dx,x,a \cos (x)\right )}{a^5}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {\left (a^2-b^2\right )^2}{(b+x)^5}-\frac {4 b \left (-a^2+b^2\right )}{(b+x)^4}-\frac {2 \left (a^2-3 b^2\right )}{(b+x)^3}-\frac {4 b}{(b+x)^2}+\frac {1}{b+x}\right ) \, dx,x,a \cos (x)\right )}{a^5}\\ &=\frac {\left (a^2-b^2\right )^2}{4 a^5 (b+a \cos (x))^4}+\frac {4 b \left (a^2-b^2\right )}{3 a^5 (b+a \cos (x))^3}-\frac {a^2-3 b^2}{a^5 (b+a \cos (x))^2}-\frac {4 b}{a^5 (b+a \cos (x))}-\frac {\log (b+a \cos (x))}{a^5}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 138, normalized size = 1.38 \[ -\frac {12 a^4 \cos ^4(x) \log (a \cos (x)+b)-3 a^4+48 a^3 b \cos ^3(x) (\log (a \cos (x)+b)+1)+12 a^2 \cos ^2(x) \left (a^2+6 b^2 \log (a \cos (x)+b)+9 b^2\right )+8 a b \cos (x) \left (a^2+6 b^2 \log (a \cos (x)+b)+11 b^2\right )+2 a^2 b^2+12 b^4 \log (a \cos (x)+b)+25 b^4}{12 a^5 (a \cos (x)+b)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cot[x] + b*Csc[x])^(-5),x]

[Out]

-1/12*(-3*a^4 + 2*a^2*b^2 + 25*b^4 + 12*b^4*Log[b + a*Cos[x]] + 12*a^4*Cos[x]^4*Log[b + a*Cos[x]] + 48*a^3*b*C
os[x]^3*(1 + Log[b + a*Cos[x]]) + 12*a^2*Cos[x]^2*(a^2 + 9*b^2 + 6*b^2*Log[b + a*Cos[x]]) + 8*a*b*Cos[x]*(a^2
+ 11*b^2 + 6*b^2*Log[b + a*Cos[x]]))/(a^5*(b + a*Cos[x])^4)

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fricas [A]  time = 2.03, size = 166, normalized size = 1.66 \[ -\frac {48 \, a^{3} b \cos \relax (x)^{3} - 3 \, a^{4} + 2 \, a^{2} b^{2} + 25 \, b^{4} + 12 \, {\left (a^{4} + 9 \, a^{2} b^{2}\right )} \cos \relax (x)^{2} + 8 \, {\left (a^{3} b + 11 \, a b^{3}\right )} \cos \relax (x) + 12 \, {\left (a^{4} \cos \relax (x)^{4} + 4 \, a^{3} b \cos \relax (x)^{3} + 6 \, a^{2} b^{2} \cos \relax (x)^{2} + 4 \, a b^{3} \cos \relax (x) + b^{4}\right )} \log \left (a \cos \relax (x) + b\right )}{12 \, {\left (a^{9} \cos \relax (x)^{4} + 4 \, a^{8} b \cos \relax (x)^{3} + 6 \, a^{7} b^{2} \cos \relax (x)^{2} + 4 \, a^{6} b^{3} \cos \relax (x) + a^{5} b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^5,x, algorithm="fricas")

[Out]

-1/12*(48*a^3*b*cos(x)^3 - 3*a^4 + 2*a^2*b^2 + 25*b^4 + 12*(a^4 + 9*a^2*b^2)*cos(x)^2 + 8*(a^3*b + 11*a*b^3)*c
os(x) + 12*(a^4*cos(x)^4 + 4*a^3*b*cos(x)^3 + 6*a^2*b^2*cos(x)^2 + 4*a*b^3*cos(x) + b^4)*log(a*cos(x) + b))/(a
^9*cos(x)^4 + 4*a^8*b*cos(x)^3 + 6*a^7*b^2*cos(x)^2 + 4*a^6*b^3*cos(x) + a^5*b^4)

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giac [A]  time = 0.13, size = 93, normalized size = 0.93 \[ -\frac {\log \left ({\left | a \cos \relax (x) + b \right |}\right )}{a^{5}} - \frac {48 \, a^{2} b \cos \relax (x)^{3} + 12 \, {\left (a^{3} + 9 \, a b^{2}\right )} \cos \relax (x)^{2} + 8 \, {\left (a^{2} b + 11 \, b^{3}\right )} \cos \relax (x) - \frac {3 \, a^{4} - 2 \, a^{2} b^{2} - 25 \, b^{4}}{a}}{12 \, {\left (a \cos \relax (x) + b\right )}^{4} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^5,x, algorithm="giac")

[Out]

-log(abs(a*cos(x) + b))/a^5 - 1/12*(48*a^2*b*cos(x)^3 + 12*(a^3 + 9*a*b^2)*cos(x)^2 + 8*(a^2*b + 11*b^3)*cos(x
) - (3*a^4 - 2*a^2*b^2 - 25*b^4)/a)/((a*cos(x) + b)^4*a^4)

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maple [A]  time = 0.13, size = 132, normalized size = 1.32 \[ \frac {4 b}{3 a^{3} \left (b +a \cos \relax (x )\right )^{3}}-\frac {4 b^{3}}{3 a^{5} \left (b +a \cos \relax (x )\right )^{3}}-\frac {4 b}{a^{5} \left (b +a \cos \relax (x )\right )}-\frac {\ln \left (b +a \cos \relax (x )\right )}{a^{5}}-\frac {1}{a^{3} \left (b +a \cos \relax (x )\right )^{2}}+\frac {3 b^{2}}{a^{5} \left (b +a \cos \relax (x )\right )^{2}}+\frac {1}{4 a \left (b +a \cos \relax (x )\right )^{4}}-\frac {b^{2}}{2 a^{3} \left (b +a \cos \relax (x )\right )^{4}}+\frac {b^{4}}{4 a^{5} \left (b +a \cos \relax (x )\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cot(x)+b*csc(x))^5,x)

[Out]

4/3*b/a^3/(b+a*cos(x))^3-4/3*b^3/a^5/(b+a*cos(x))^3-4*b/a^5/(b+a*cos(x))-ln(b+a*cos(x))/a^5-1/a^3/(b+a*cos(x))
^2+3/a^5/(b+a*cos(x))^2*b^2+1/4/a/(b+a*cos(x))^4-1/2/a^3/(b+a*cos(x))^4*b^2+1/4/a^5/(b+a*cos(x))^4*b^4

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maxima [B]  time = 0.48, size = 497, normalized size = 4.97 \[ -\frac {2 \, {\left (5 \, a^{4} b + 10 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 6 \, a b^{4} - 3 \, b^{5} + \frac {{\left (3 \, a^{5} - 17 \, a^{4} b - 6 \, a^{3} b^{2} + 26 \, a^{2} b^{3} + 3 \, a b^{4} - 9 \, b^{5}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {3 \, {\left (4 \, a^{5} - 13 \, a^{4} b + 12 \, a^{3} b^{2} + 2 \, a^{2} b^{3} - 8 \, a b^{4} + 3 \, b^{5}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {3 \, {\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}}\right )}}{3 \, {\left (a^{10} + 2 \, a^{9} b - a^{8} b^{2} - 4 \, a^{7} b^{3} - a^{6} b^{4} + 2 \, a^{5} b^{5} + a^{4} b^{6} - \frac {4 \, {\left (a^{10} - 3 \, a^{8} b^{2} + 3 \, a^{6} b^{4} - a^{4} b^{6}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {6 \, {\left (a^{10} - 2 \, a^{9} b - a^{8} b^{2} + 4 \, a^{7} b^{3} - a^{6} b^{4} - 2 \, a^{5} b^{5} + a^{4} b^{6}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {4 \, {\left (a^{10} - 4 \, a^{9} b + 5 \, a^{8} b^{2} - 5 \, a^{6} b^{4} + 4 \, a^{5} b^{5} - a^{4} b^{6}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}} + \frac {{\left (a^{10} - 6 \, a^{9} b + 15 \, a^{8} b^{2} - 20 \, a^{7} b^{3} + 15 \, a^{6} b^{4} - 6 \, a^{5} b^{5} + a^{4} b^{6}\right )} \sin \relax (x)^{8}}{{\left (\cos \relax (x) + 1\right )}^{8}}\right )}} - \frac {\log \left (a + b - \frac {{\left (a - b\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{5}} + \frac {\log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^5,x, algorithm="maxima")

[Out]

-2/3*(5*a^4*b + 10*a^3*b^2 + 2*a^2*b^3 - 6*a*b^4 - 3*b^5 + (3*a^5 - 17*a^4*b - 6*a^3*b^2 + 26*a^2*b^3 + 3*a*b^
4 - 9*b^5)*sin(x)^2/(cos(x) + 1)^2 - 3*(4*a^5 - 13*a^4*b + 12*a^3*b^2 + 2*a^2*b^3 - 8*a*b^4 + 3*b^5)*sin(x)^4/
(cos(x) + 1)^4 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sin(x)^6/(cos(x) + 1)^6)/(a^10 +
2*a^9*b - a^8*b^2 - 4*a^7*b^3 - a^6*b^4 + 2*a^5*b^5 + a^4*b^6 - 4*(a^10 - 3*a^8*b^2 + 3*a^6*b^4 - a^4*b^6)*sin
(x)^2/(cos(x) + 1)^2 + 6*(a^10 - 2*a^9*b - a^8*b^2 + 4*a^7*b^3 - a^6*b^4 - 2*a^5*b^5 + a^4*b^6)*sin(x)^4/(cos(
x) + 1)^4 - 4*(a^10 - 4*a^9*b + 5*a^8*b^2 - 5*a^6*b^4 + 4*a^5*b^5 - a^4*b^6)*sin(x)^6/(cos(x) + 1)^6 + (a^10 -
6*a^9*b + 15*a^8*b^2 - 20*a^7*b^3 + 15*a^6*b^4 - 6*a^5*b^5 + a^4*b^6)*sin(x)^8/(cos(x) + 1)^8) - log(a + b -
(a - b)*sin(x)^2/(cos(x) + 1)^2)/a^5 + log(sin(x)^2/(cos(x) + 1)^2 + 1)/a^5

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mupad [B]  time = 3.68, size = 538, normalized size = 5.38 \[ \frac {2\,\mathrm {atanh}\left (\frac {32\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{\frac {32\,b^3}{a^3}-\frac {32\,b^2}{a^2}-\frac {32\,b}{a}+\frac {32\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a}-\frac {64\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}+\frac {32\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^3}+32}-\frac {64\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{32\,a-32\,b+32\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-\frac {32\,b^2}{a}+\frac {32\,b^3}{a^2}-\frac {64\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a}+\frac {32\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}}+\frac {32\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{32\,a^2-32\,a\,b-32\,b^2-64\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\frac {32\,b^3}{a}+\frac {32\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a}+32\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}\right )}{a^5}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6\,\left (a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{a^4}+\frac {2\,\left (5\,a^4\,b+10\,a^3\,b^2+2\,a^2\,b^3-6\,a\,b^4-3\,b^5\right )}{3\,a^4\,{\left (a-b\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (-4\,a^3+5\,a^2\,b+2\,a\,b^2-3\,b^3\right )}{a^4}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (3\,a^4-14\,a^3\,b-20\,a^2\,b^2+6\,a\,b^3+9\,b^4\right )}{3\,a^4\,\left (a-b\right )}}{4\,a\,b^3+4\,a^3\,b+{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (6\,a^4-12\,a^2\,b^2+6\,b^4\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (-4\,a^4-8\,a^3\,b+8\,a\,b^3+4\,b^4\right )-{\mathrm {tan}\left (\frac {x}{2}\right )}^6\,\left (4\,a^4-8\,a^3\,b+8\,a\,b^3-4\,b^4\right )+a^4+b^4+{\mathrm {tan}\left (\frac {x}{2}\right )}^8\,\left (a^4-4\,a^3\,b+6\,a^2\,b^2-4\,a\,b^3+b^4\right )+6\,a^2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/sin(x) + a*cot(x))^5,x)

[Out]

(2*atanh((32*tan(x/2)^2)/((32*b^3)/a^3 - (32*b^2)/a^2 - (32*b)/a + (32*b*tan(x/2)^2)/a - (64*b^2*tan(x/2)^2)/a
^2 + (32*b^3*tan(x/2)^2)/a^3 + 32) - (64*b*tan(x/2)^2)/(32*a - 32*b + 32*b*tan(x/2)^2 - (32*b^2)/a + (32*b^3)/
a^2 - (64*b^2*tan(x/2)^2)/a + (32*b^3*tan(x/2)^2)/a^2) + (32*b^2*tan(x/2)^2)/(32*a^2 - 32*a*b - 32*b^2 - 64*b^
2*tan(x/2)^2 + (32*b^3)/a + (32*b^3*tan(x/2)^2)/a + 32*a*b*tan(x/2)^2)))/a^5 - ((2*tan(x/2)^6*(3*a*b^2 - 3*a^2
*b + a^3 - b^3))/a^4 + (2*(5*a^4*b - 6*a*b^4 - 3*b^5 + 2*a^2*b^3 + 10*a^3*b^2))/(3*a^4*(a - b)^2) + (2*tan(x/2
)^4*(2*a*b^2 + 5*a^2*b - 4*a^3 - 3*b^3))/a^4 + (2*tan(x/2)^2*(6*a*b^3 - 14*a^3*b + 3*a^4 + 9*b^4 - 20*a^2*b^2)
)/(3*a^4*(a - b)))/(4*a*b^3 + 4*a^3*b + tan(x/2)^4*(6*a^4 + 6*b^4 - 12*a^2*b^2) + tan(x/2)^2*(8*a*b^3 - 8*a^3*
b - 4*a^4 + 4*b^4) - tan(x/2)^6*(8*a*b^3 - 8*a^3*b + 4*a^4 - 4*b^4) + a^4 + b^4 + tan(x/2)^8*(a^4 - 4*a^3*b -
4*a*b^3 + b^4 + 6*a^2*b^2) + 6*a^2*b^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))**5,x)

[Out]

Timed out

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